The differentiable functions $x$ and $y$ are related by the following equation: $\dfrac{1}{y}=8-x$ Also, $\dfrac{dy}{dt}=-0.5$. Find $\dfrac{dx}{dt}$ when $y=0.2$.
Answer: Let's start by differentiating the equation $\dfrac{1}{y}=8-x$ with respect to $t$. $\begin{aligned} \dfrac{1}{y}&=8-x \\\\ -\dfrac{1}{y^2}\cdot\dfrac{dy}{dt}&=-\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dy}{dt}=-0.5$, and we want to find $\dfrac{dx}{dt}$ when $y=0.2$. Let's plug ${y=0.2}$ and ${\dfrac{dy}{dt}=-0.5}$ into the equation we obtained: $\begin{aligned} -\dfrac{1}{{y}^2}\cdot{\dfrac{dy}{dt}}&=-\dfrac{dx}{dt} \\\\ -\dfrac{1}{{0.2}^2}({-0.5})&=-\dfrac{dx}{dt} \\\\ -12.5&=\dfrac{dx}{dt} \end{aligned}$ In conclusion, when $y=0.2$, the value of $\dfrac{dx}{dt}$ is $-12.5$.